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Question 2

If the set of all solutions of $$|x^2 + x - 9| = |x| + |x^2 - 9|$$ is $$[\alpha, \beta] \cup [\gamma, \infty)$$, then $$(\alpha^2 + \beta^2 + \gamma^2)$$ is equal to :

The given equation is
$$|x^{2}+x-9| = |x| + |x^{2}-9|$$

The expressions inside the absolute values change sign at the real zeros of
$$x = 0,\quad x = \pm 3,\quad x = \frac{-1\pm\sqrt{37}}{2}\;(\approx -3.541,\; 2.541).$$
Hence analyse the equation in the six intervals determined by these points.

Case 1: $$x\lt\frac{-1-\sqrt{37}}{2}\;(\;x\lt-3.541\;).$$

Here $$x&lt0,\;x^{2}-9&gt0,\;x^{2}+x-9&gt0$$ giving

$$x^{2}+x-9 = -x + (x^{2}-9)\implies x=0$$

No value in this interval satisfies the equation.

Case 2: $$\frac{-1-\sqrt{37}}{2}\lt x\lt -3.$$

Now $$x&lt0,\;x^{2}-9&gt0,\;x^{2}+x-9&lt0$$ so

$${-(x^{2}+x-9)} = -x + (x^{2}-9)$$ $$\Longrightarrow 2x^{2}=18\Longrightarrow x=\pm3.$$

Neither value lies in this open interval, hence no solution here.

Case 3: $$-3\lt x\lt 0.$$

Here $$x&lt0,\;x^{2}-9&lt0,\;x^{2}+x-9&lt0$$ giving

$${-(x^{2}+x-9)} = -x + (-(x^{2}-9))$$ $$\Longrightarrow -x^{2}-x+9 = -x^{2}-x+9,$$

which is an identity. Therefore every $$x$$ in $$(-3,0)$$ is a solution.

Case 4: $$0\lt x\lt\frac{-1+\sqrt{37}}{2}\;(\;0\lt x\lt 2.541\;).$$

Now $$x&gt0,\;x^{2}-9&lt0,\;x^{2}+x-9&lt0$$ so

$${-(x^{2}+x-9)} = x + (-(x^{2}-9))$$ $$\Longrightarrow -x^{2}-x+9 = -x^{2}+x+9$$ $$\Longrightarrow x=0.$$

Again, no interior point meets the requirement; only the boundary point $$x=0$$ works.

Case 5: $$\frac{-1+\sqrt{37}}{2}\lt x\lt 3.$$

Here $$x&gt0,\;x^{2}-9&lt0,\;x^{2}+x-9&gt0$$ so

$$x^{2}+x-9 = x + (-(x^{2}-9))$$ $$\Longrightarrow 2x^{2}=18\Longrightarrow x=\pm3.$$

The only admissible value would be $$x=3,$$ but $$x=3$$ is not inside this open interval.

Case 6: $$x\ge 3.$$

Now all three quantities are non-negative, so

$$x^{2}+x-9 = x + (x^{2}-9),$$

which is always true. Hence every $$x\ge3$$ satisfies the equation.

Finally test the remaining critical points:

• $$x=-3:\;|(-3)^2-3-9| = 3,\;|{-3}|+|9-9| = 3\Rightarrow$$ true.
• $$x=0:\;|0-9| = 9,\;0+|{-9}| = 9\Rightarrow$$ true.
• $$x=3:\;|9+3-9| = 3,\;3+0 = 3\Rightarrow$$ true.

Therefore the complete solution set is
$$[-3,0]\cup[3,\infty).$$

Identifying $$\alpha=-3,\;\beta=0,\;\gamma=3,$$ we have

$$\alpha^{2}+\beta^{2}+\gamma^{2}=(-3)^{2}+0^{2}+3^{2}=9+0+9=18.$$

Option B which is: 18

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